May 6, 2015 · Which method do you use to factor #3x(x-1)+4(x-1) #? What are the factors of #12x^3+12x^2+3x#? How do you find the two numbers by using the factoring method, if one number is seven more than...

Dec 15, 2017 · Expanding upon prior answer: I want to expand upon an idea expressed in the prior answer The idea of: (x^n - 1)/(x-1) = sum_(r=1) ^n x^(n-r) or not in sigma notation: (x^n -1 )/(x-1) = x^(n-1) + x^(n-2) + ... + x + 1 We can prove this via induction: Basis case : => n = 1 LHS: (x^1-1)/(x-1) = 1 RHS: x^(1-1) = x^0 = 1 Hence basis case holds Induction: Assume n=k holds: (x^k - 1)/(x-1) = sum_(r=1 ...

Apr 13, 2018 · 87526 views around the world You can reuse this answer ...

Jan 3, 2016 · Which method do you use to factor #3x(x-1)+4(x-1) #? What are the factors of #12x^3+12x^2+3x#? How do you find the two numbers by using the factoring method, if one number is seven more than...

Nov 30, 2016 · 67059 views around the world You can reuse this answer ...

Aug 16, 2016 · Using synthetic division and the fact that #x=-1# is obviously a solution we find that we can expand this to: #(x+1)(x^2-x+1) = 0# In order to have LHS=RHS need one of the brackets to be equal to zero, ie

Apr 24, 2017 · See below. For my approach, I will be using a graphical interpretation. You can rewrite the equation as x^3 - x - 1 = 0 as the first step. Then graph the following: f(x) = x^3 - x - 1. graph{x^3 - x - 1 [-10, 10, -5, 5]} Click on where the graph intersects the x-axis. This point should be (1.325, 0). Therefore, the answer is x = 1.325.

Apr 11, 2016 · (x^2 - x + 1) Apply the identity: a^3 + b^3 = (a + b)(a^2 - ab + b^2) x^3 + 1 = (x + 1)(x^2 - x + 1) Therefor: (x^3 + 1)/(x + 1) = (x^2 - x + 1)

Feb 25, 2018 · What is the integration of 1/x3? Calculus. 1 Answer Pranav M. Feb 25, 2018 #-1/(2x^2)+c#. Explanation: to ...

Jan 21, 2017 · Use partial fraction expansion and then integrate Partial fraction expansion of 1/(x^3 +1): -1 is a root ...